1. Problem: A couple plans to have three children. Find the probabilities for different gender combinations. 2. Formula: For independent events, probability of a specific outcome is given by $$P = \frac{\text{number of favorable outcomes}}{\text{total possible outcomes}}$$. 3. Important rules: - Each child is equally likely to be a boy or a girl, so $P(\text{boy}) = P(\text{girl}) = \frac{1}{2}$. - The total number of outcomes for three children is $2^3 = 8$. 4. a. Probability all children are girls: - Only one outcome: GGG. - So, $$P(\text{all girls}) = \frac{1}{8}$$. 5. b. Probability exactly two children are boys: - Possible outcomes with exactly two boys: B B G, B G B, G B B. - Number of favorable outcomes = 3. - So, $$P(\text{2 boys}) = \frac{3}{8}$$. 6. c. Probability at least 2 children are girls: - At least 2 girls means 2 or 3 girls. - Outcomes with 2 girls: G G B, G B G, B G G (3 outcomes). - Outcome with 3 girls: G G G (1 outcome). - Total favorable outcomes = 3 + 1 = 4. - So, $$P(\text{at least 2 girls}) = \frac{4}{8} = \frac{1}{2}$$. 7. d. Probability couple will have 2 boys at most: - "2 boys at most" means 0, 1, or 2 boys. - Number of outcomes with 0 boys (all girls): 1. - Number of outcomes with 1 boy: B G G, G B G, G G B (3 outcomes). - Number of outcomes with 2 boys: 3 (from part b). - Total favorable outcomes = 1 + 3 + 3 = 7. - So, $$P(\text{2 boys at most}) = \frac{7}{8}$$. 8. Problem: Rolling a pair of dice once, find the probability that the sum is a specific number. 9. Total possible outcomes when rolling two dice: $6 \times 6 = 36$. 10. a. Probability sum is 5: - Possible pairs: (1,4), (2,3), (3,2), (4,1). - Number of favorable outcomes = 4. - So, $$P(\text{sum}=5) = \frac{4}{36} = \frac{1}{9}$$. 11. b. Probability sum is 9: - Possible pairs: (3,6), (4,5), (5,4), (6,3). - Number of favorable outcomes = 4. - So, $$P(\text{sum}=9) = \frac{4}{36} = \frac{1}{9}$$.